MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

To phrase the question in a concrete way, I read in a paper:

The super Poincare subalgebra of osp(6,2|4) has bosonic part $so(5,1) \oplus usp(4) \simeq so(5,1) \oplus so(5)$.

It's hard to unpack this sentence without knowing the objects:

  • Can the orthosymplectic group osp(6,2|4) be defined as a group of matrices acting on a super-vector space?
  • Can someone explain this decomposition into its bosonic and fermionic parts?
  • What is the (super) Poincare sub-algebra? Why is usp(4) the same is so(5)?

For a math-physics dictionary: "super" means "Z2-graded" while bosonic means "0-grade" and fermionic means "1-grade".

share|cite|improve this question

The way I would understand it that $osp(6,2|4)$ is the group of linear tansformations of a real super vector space with a non-degenerate symmetric inner product. The even (bosonic) vector space has dimension 8 and the inner product is symmetric with signature $(6,2)$ the odd (fermionic) part has dimension 4 and a symplectic form.

The even part is then the product of the groups of these two vector spaces, namely $o(6,2)$ and $sp(4)$. There is an isomophism of rank two Lie algebras $sp(4)\cong so(5)$; to see this note that the spin representation has dimension 4 and has an invariant symplectic form.

I realise you have $so(5,1)$ where I have $so(6,2)$. I don't know what is going on here but $so(6,2)$ is the group of conformal transformations of $R^{5,1}$.

share|cite|improve this answer
I imagine that your failure to use the phrase "Poincare subalgebra" cancels your other discrepancy (the last paragraph), but I don't know how. The usual Poincare algebra is the semidirect product of the $so(3,1)$ with $R^{3,1}$. (A super Poincare algebra has that as the bosonic part.) They're probably embedding some generalization of it in a larger group. It seems weird that they could shove the normal subgroup into the fermionic part. Maybe they left out "semisimplification"? – Ben Wieland May 27 '10 at 22:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.